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Is lift less in a steeper descent than in a shallower descent, and if so, by how much?
Is lift less in a steeper descent than in a shallower descent, and if so, by how much?
For example, if we change our ratio of horizontal distance travelled to vertical distance travelled (through the airmass) from 10:1 (5.7 degree descent angle) to 5:1 (11.3 degree descent angle)-- which, if we hold the airspeed constant, very nearly doubles the sink rate-- have we decreased the lift force, and by how much?
For relatively shallow descent angles (say up to 15 or 20 degrees), does it really make sense to say that if we leave thrust constant, we can increase the descent rate and make the glide path steeper "by reducing lift", meaning the actual lift force as we would measure in Newtons or pounds?  Or does the idea of descending by "reducing lift" have some other interpretation that is more accurate or meaningful?
Does the answer depend on whether we are increasing the descent rate and making the glide path steeper by opening spoilers, or by deploying flaps, or by reducing power, or by just putting the nose down and allowing the airspeed to increase, assuming we end up with the same change in glide path in every case?
For example, what really happens, in terms of the lift force, when we open spoilers to "dump lift" and make our descent angle steeper, while holding airspeed constant, without changing thrust?  (Assume that the "spoilers" really do decrease the lift coefficient, at the instant that they are deployed.)  After we are established on the new flight path, has the lift force really decreased, and if so, by how much?  (Specifically, for the change in glide path described above.)
Assume the conventional definitions of lift and drag-- i.e. lift acts perpendicular to the flight path, and drag acts parallel to the flight path.  Start by assuming that the thrust line is parallel to the flight path; feel free to elaborate on other variations if you feel that is warranted.  Assume we are always talking about the glide path and sink rate through the airmass, i.e. what we'd achieve over the ground if there were no wind-- this is not a "trick question" about effects due to wind.

This question has a similar title https://aviation.stackexchange.com/questions/56718/is-lift-force-less-than-weight-during-a-descent but it is really about a misconception on a table, plus a few other questions thrown in; the body of the question is not a duplicate.

This question https://aviation.stackexchange.com/q/56254/34686, when correctly understood, has similarities to the present question, but the question is posed in a very different way and invites some very different interpretations

Maybe I don't know enough, but this seems to be very broad. There are a lot of question marks in there. Maybe try narrowing the focus some? Or at least the question mark count...

Well, I could truncate it after the second or third paragraph--

@FreeMan -- better?

I think this question can be answered as long as no spoilers/flaps are extended (i.e. L/D is the same). But once you change the configuration (spoilers/flaps), you need to specify how that changes the L/D.

@Bianfable -- my opinion is otherwise, I think all the needed constraints are given--

@Bianfable -- because the thing is, every change you make, including a change in angle-of-attack, is going to change the L/D.  So there's no way to eliminate that.  Unless you say that all you are doing is reducing power (thrust), and nothing else.  But that's not the question I want to ask.

@Bianfable -- although actually, that was a part of my question before I simplified it in response to Freeman's comment--

I put the flaps etc back in, but not as a bunch of separate paragraphs, keeping it shorter.

Yes.
in an unaccelerated descent - or ascent -, lift - the force acting perpendicular to the direction of motion - scales with the cosine of the angle of flight from horizontal. Hence, an aircraft in a 90 degree dive will have no lift.
This is independent of airspeed or the method by which lift reduction is achieved.
Here's a graph of lift Vs glide ratio

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Um... lift is a force acting perpendicular to the relative airlow, therefore it is still present in a vertical dive, but now it doesn't oppose gravity .....

@Raffles No. If it were, the aircraft would accelerate towards the horizon and pitch upeas a result.

you need to specify how that changes Lift/Drag ... Bianfable

That's the answer.
Lift requirement in descent is defined the same way as in a climb:

Cosine angle(from horizon) x weight

The rest of the vertical lift requirement is from the the vertical drag component.  (In a climb it is from the vertical thrust component).
When you dump spoilers and flaps, this drastically increases the drag, requiring a higher angle of descent to maintain airspeed.
Flaps increase lift coeficient and increase drag allowing for a slower, steeper descent.
Spoilers decrease lift coefficient and increase drag, allowing for a faster, steeper descent.  This very valuable to a glider trying to land under windy conditions.
But the lift requirement for stable, linear flight is determined by weight x cos angle to horizon, no matter what combination of drag, lift coefficient, and airspeed one uses.

@quiet flyer that's ok, I think you are correct that "residual thrust" does not make it more complicated, it is merely subtracted from drag in the glide ratio: D-T = Net Drag. A little thrust flattens the glide (same effect as reducing drag), more and D-T=0  D=T.(level flight).

Re "Spoilers decrease lift coefficient and increase drag, allowing for a faster, steeper descent. This very valuable to a glider trying to land under windy conditions." -- food for thought-- I'd argue that in actual practice, a glider pilot can always increase speed via the elevator control, and I'd suggest that flaps (especially full-span, deflected near 90 degrees) do a better job of increasing the sink rate at high airspeeds than do spoilers (at least so long as we are still talking about glide angles of 20 degrees or less).

(But flaps may have a redline for deployment lower than max aircraft redline.)  So why do gliders have spoilers instead of flaps?  I think there's a great ASE question in there, someone should ask it.  Possible answers may include: better control of glide path at relatively low airspeed (e.g during the "float" before touchdown), ease of rapid deployment, ability to create a design that may be deployed at all airspeeds up to redline and prevents the aircraft from exceding redline even when lost in a cloud etc w/ nose straight down ...

(ctd) , ability to create a design that may be deployed at all airspeeds up to redline and prevents the aircraft from exceding redline even when lost in a cloud etc w/ nose straight down (the latter point is not a universal feature of all glider spoiler designs.)

@quiet flyer gliders have spoilers to essentially to "shrink the wing", this allows the plane to land at higher airspeed, perhaps less ground effect with spoilers.  But yes, massively deflected flaps force a steeper approach and rapidly drop airspeed after the flare.  Notice airliners do both, and reverse thrusters too!

Other answers have already pointed out that

In a steady-state descent, for the simple case where the thrust vector is considered to act parallel to the flight path, lift =weight * cosine (descent angle).

Therefore, the larger the descent angle, the smaller the lift vector must be.

However, for reasonably small descent angles-- say 20 degrees or less-- this effect is very small, and (if we assume the thrust line acts parallel to the flight path), lift remains very nearly equal to weight.

For example, at a 10 degree descent angle, which in the context  of an instrument approach would be considered absurdly steep, lift is still about 98% of weight. Going from a 10:1 descent ratio (5.7 degree descent angle) to a 5:1 descent ratio (11.3 degree descent angle), we very nearly double the sink rate (if airspeed is constant), but the lift vector only changes from 99.5% of weight to 98.1% of weight, a change of about 1.5% (i.e. the ratio between the two numbers is about 1.015).
Things are different at steeper descent angles, especially well over 30 degrees, where the (drag minus thrust) vector carries a significant part of the aircraft weight-- all of it, in the extreme case of a vertical dive.  At steep descent angles, further increasing the descent angle does significantly reduce the magnitude of the lift force.  (Also, at very steep descent angles-- above 45 degrees-- increasing the magnitude of the (drag minus thrust) vector actually decreases the sink rate, rather than increasing it.)1
As long as the descent angle is small, when we say we increase the descent angle and sink rate "by reducing lift" -- e.g. by opening spoilers-- it is technically true that the lift force is slightly reduced, but if we are envisioning a significant effect, what we really mean must be something else entirely.
In some cases, we might really mean that we are reducing the lift coefficient.  The descent angle is determined by the ratio of Lift to (Drag minus Thrust), which in the power-off case, is exactly equal to the ratio of the lift coefficient to the drag coefficient.  This gives some insight into how reducing the lift coefficient as well as increasing the drag coefficient, can help us to descend more steeply.
However, we have to be careful about how we apply this idea as well, for at least a couple of reasons.  One, because it gives no insight into how we can descend by deploying flaps, or by reducing thrust.  And two,  because it gives no insight into any case where we are holding the airspeed constant.  Whenever the airspeed is constant, the lift coefficient and the lift force are directly related, so the lift coefficient must be directly related to weight * cosine (descent angle).  For shallow descent angles, we've already noted that the value of this expression only decreases slightly even when we make a large increase in the descent angle and sink rate.
So what is really going on when we increase our descent angle and sink rate by deploying spoilers, or deploying flaps, or reducing power, while holding airspeed constant?
If we deploy the flaps, this will have the initial effect of increasing the lift coefficient as well as the drag coefficient.  If we don't want to lose airspeed, we'll need to decrease the angle-of-attack of the wing to return to (very nearly) the original lift coefficient.   Decreasing the angle-of-attack in this manner slightly reduces the net increase in drag coefficient that would otherwise be created by deploying the flaps, but the drag coefficient is still higher than it was before we deployed the flaps. Unless the airspeed is quite low, we have more drag force and a higher descent angle and a higher sink rate when the flaps are down than when they are up.  The higher the airspeed, the more pronounced this effect.  When the flaps are deployed to a very large deflection, such as 45 degrees or more, lowering the nose to increase the airspeed can be a very effective way to create a very large increase in the descent angle.  Another way to say this is to note that deploying the flaps creates a very large downward shift (toward a higher sink rate) in the far right-hand portion of the "polar curve" of airspeed versus vertical speed.
If we increase the descent angle by opening the spoilers, this will have the initial effect of reducing the lift coefficient as well as increasing the drag coefficient.  If we don't want to gain airspeed, we'll need to increase the angle-of-attack of the wing to return to (very nearly) the original lift coefficient.  The increase in angle-of-attack also further increases the drag coefficient.  The net result on the sink rate and descent angle is somewhat similar to if we had deployed the flaps, except that there is no decrease in stall speed.  Rather, the stall speed is raised.
(Note that increases or decreases in angle-of-attack (as measured at the fuselage or the unflapped part of the wing) do not directly correspond to identical increases or decreases in pitch attitude, especially if we're simultaneously making a change in configuration, which also affects the flight path.  Note also that for simplicity this answer assumes wing area is constant-- the increased wing area due to deploying Fowler flaps, for example, is just treated as an increase in the lift and drag coefficients.  Note also that throughout this answer, for simplicity we treat the thrust  vector as acting parallel to the flight path through the airmass.)
At the end of the day, while flying at any given constant airspeed, for shallow descent angles, we increase our descent angle and our descent rate by reducing thrust or by adding drag.  The closed vector diagram of Weight, Lift, and (Drag - Thrust) does force a small decrease in the value of the Lift vector whenever we increase the descent angle, but for shallow descent angles, this effect is very small.
Note that if we are operating on the "front side" of the thrust-required curve (i.e. not at extremely low airspeeds), simply moving the control stick forward is another way to add drag, even though the drag coefficient is actually reduced by the decrease in angle-of-attack.  What we're really doing here is reducing both the lift and drag coefficients, which leads to a higher airspeed, which (for small descent angles) restores the lift force to (very nearly) its original value, but at cost of a net increase in drag.  So the descent angle and sink rate are both increased.
Footnotes:

Equation needed, to prove that this statement is true, for a given constraint (e.g. constant lift coefficient?  Or constant descent angle?)

Related ASE questions--
Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
(this answer contains vector diagrams relevant to the present answer, and also includes (near the end) a discussion of effects arising from the thrust line not being parallel to the flight path)
'Gravitational' power vs. engine power -- the variation in the size of the Lift vector as the flight path changes from horizontal to descending comes into play here, exactly as it does in the present answer
Does lift equal weight in a climb?

Sorry, composing on phone is challenging, finished now-- !

Re footnote: I think I worked out the math on this once before but don't have it handy at the moment--


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In this F-104 V-n diagram, why does the stall speed (in terms of IAS) decrease with altitude in some parts of the flight envelope?

(This question has been edited to reflect a change in perspective: based on answers to the related question What is causing these "corners" on this F-104 V-n diagram?, I now believe that the curves on the upper left edge of the envelope are intended to represent the stall speed throughout their entirety, both above and below the distinct "corners".  If the aircraft is being limited by something other than the actual stall along some portions of these curves, then it would be helpful to clarify and explain that in an answer either to the present question or the one linked above.)
In this V-n diagram for an F-104 (source: this Wikipedia link), why does the stall speed at any given G-loading (in terms of IAS) decrease with altitude-- at least in the part of the envelope that is below the sharp "corners"?
And why is the opposite is true for data points above the "corners"?
For example-- (numbers are approximate)--
Below the "corners"--
20,000', 5G, 415 knots IAS, versus 30,000', 5G, 375 knots IAS
30,000', 3G, 330 knots IAS, versus 40,000', 3G, 295 knots IAS
60,000', 1G, 175 knots IAS, versus 70,000', 1G, 140 knots IAS
Above the "corners"--
30,000', 5G, 375 knots IAS, versus 40,000', 5G, 475 knots IAS
And why do the curves cross each other?
(For example, at about 350 KIAS and 3.5G, the 40,000' curve crosses the 30,000' curve.)
Related ASE questions and answers-- but some of which seem to suggest the opposite relationship should be the case (at least for the "below the corners" cases)--
Why does indicated stall speed change? (see question and all answers)
What causes a slight increases in indicated stall speed with altitude? (see question and all answers)
How does the IAS stall speed vary with increasing altitude? (see question and all answers)

@ymb1-- lots of tags deleted there (edit #4), was that all really necessary?  Among other things, stall speed is arguably not unrelated to aircraft performance, and physics is a pretty broad brush that would seem to include this question.

After much magnification, it seems each altitude has only one step. Notice S.L. and 20,000 feet do not have a step.  The get progressively lower (lower IAS) with altitude.

@RobertDiGiovanni -- yes, agreed

As an airplane approaches Mach 1, all pressure changes grow with the Prandtl-Glauert factor of $\frac{1}{\sqrt{1-Ma^2}}$. Therefore, the lift curve slope increases so the wing produces more lift at the same angle of attack and dynamic pressure the closer its Mach number is to 1. On wings with thicker airfoils and higher aspect ratio the maximum lift coefficient drops in the transsonic region; however, thin wings of low aspect ratios are unaffected by this.
NACA TN 3469 by Edward Polhamus compares wings with varying sweep angle, thickness and aspect ratio. The plot below shows that the Prandtl-Glauert effect is strongest for unswept wings and the transsonic lift bucket disappears for thin wings with an aspect ratio of 4 and lower. The F-104 had a 3.36% airfoil and an aspect ratio of 2.45.

This effect also shows up in the maximum lift; the plot below is from Sighard Hoerner's Fluid Dynamic Lift, chapter 4. It shows how the Mach number affects maximum lift: Small leading edge radii show a loss of maximum lift at moderate Mach numbers but perform well at transsonic speeds.

The indicated stall speed decreases with altitude because the lower temperature lowers the local speed of sound and brings the airplane closer to Mach 1 and thus to higher maximum lift. This effect is limited to subsonic speed close to Mach 1 which explains why those corners only show up when the indicated airspeed correlates with Mach close to 1 at that particular altitude.

Plugging in some values using a True Airspeed calculator seems to confirm Peter Kampf's conclusion that the "steps" are indeed coinciding with transonic flight at various IAS and altitudes.
There for, there is no "over lapping" of the curves, though they appear to be on the graph.
One might suggest the flattening of the curve once the aircraft goes supersonic may be related to "Mach tuck", and some sensible engineering related to airframe stress.
By design, and because of the rearward shift of the center of lift, after trimming there simply is not enough "up" stabilizer left to re-approach stall limits in supersonic flight.

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Terminology: is the "normal acceleration" (nz) considered to be 1 G, or 0, when an aircraft is at rest on the ground with fuselage horizontal?
This is meant to be a question about terminology, not a request for an explanation of the underlying physics at play.
When an airplane is in constant-speed straight-and-level upright flight with the longitudinal axis horizontal, or when an airplane is at rest on level ground with the longitudinal axis horizontal, is the "normal acceleration" (nz), which is defined as the component of the linear acceleration of an aircraft along the body Z axis, considered to be 1 G, or 0, or does it vary depending on the context?
If the latter, how so? For example, what answer would a flight test engineer give?
Bonus question-- in cases where the "normal acceleration" (nz) is considered to be 0 G rather than 1 G in constant-speed straight-and-level upright flight with the aircraft's longitudinal axis horizontal -- if any such cases exist--- then what is the "normal acceleration" (nz) considered to be in a constant-speed 60-degree-banked turn with the aircraft's longitudinal axis horizontal, with the G-meter reading 2 G's?  Is nz considered to be 1 G, or is nz considered to be 1.5 G?  In other words, have we simply shifted all the nz values downward by 1 G, or have we switched to an entirely different method of calculating nz, based on the net acceleration acting on the aircraft, rather than the felt or non-gravitational acceleration acting on the aircraft?  (For more context, see this related ASE answer.)

Not a duplicate of https://aviation.stackexchange.com/questions/95313/aircraft-load-factor-and-body-normal-acceleration , because that question is so broad (contains many questions in one) that no answer is likely to decisively answer this question.

I think the real answer is given here -- https://aviation.stackexchange.com/a/95531/34686

Highly relevant, esp comments from 10-27-22 onwards-- https://chat.stackexchange.com/rooms/140031/discussion-on-answer-by-sophit-terminology-is-the-normal-acceleration-nz-co -- (answer was apparently deleted) -- adopting the 0G convention for straight and level flight leads to real problems in other attitudes.

At zero G, you're floating (Vomit Comet ride). At 1 G, you have normal weight. At 2 G's, you seem to weigh twice as much as usual (60 degree banked turn).
The normal aviation convention is that 1 G is straight & level unaccelerated flight, same as sitting on the ground.
A typical aircraft G-meter:

Note that G's are not acceleration; they are force per unit mass. They may produce acceleration, but if the answer is expressed in G's, then the force produced by gravity counts. Even when it's being opposed by the push of the chair you're sitting in.
If the question is "what is your acceleration" (delta velocity) then the answer comes in units of meters per second-squared, and sitting in the chair on the floor, the answer is zero. If the question is, how many G's do you experience in that condition, the answer is in units of G's (or force per mass), and the answer is 1.0.  If the question is "what do you weigh," then the answer comes in units of force (i.e. Newtons). If the question is "how much mass do you have," then the answer comes in units of mass, i.e. kilograms.
If the expected answer will have units of G's, then (on earth) when sitting still, the scaler part of the answer is 1. On the moon, it would be 1/6th of a G (even tho delta V would still be zero).

To be entirely technically correct, it should be noted that the units of g and of acceleration are the same. The SI units of force per unit mass are N/kg = (kg * m/s²)/kg = m/s². That is not a coincidence: g is an acceleration: the acceleration observed by an object in free fall near the earth's surface. In special relativity there isn't even a real difference between gravity and acceleration. Conventionally of course it can make sense to make a distinction.

Re "but if the answer is expressed in G's, then the force produced by gravity counts." -- but does it really?  Sitting in chair-- chair is pushing up on you at 1 G  * mass-- hence the G-meter you are holding in your hand reads 1 G.  Even though gravity is also pulling down at 1 G * mass, producing a net force of zero.  Then the floor breaks and you are falling and gravity is still exerting a force on you, yet the G-meter reads zero.  Related: https://aviation.stackexchange.com/questions/95313/aircraft-load-factor-and-body-normal-acceleration

@quietflyer "G's" are force experienced, not force exerted. Free-falling & experiencing "0 G" in the Vomit Comet means you feel 0 G's, not that Earth's gravity no longer acts on you. If that weren't the case, we could never discuss an "zero G" situation.

Yes, but force "experienced" or "felt" by object is identical to the non-gravitational component of the net force exerted on object.  (Or your could say, the mirror image of)
That's why I've been talking about subtracting out the gravitational force from the net force.  And that's why the G-meter tells you the aerodynamic force (per unit mass) acting in the body Z axis, which is essentially the same as the lift force (per unit mass) generated by the wings.  (Except for the discrepancy due to lift vector not always being perfectly aligned with body Z axis.)

Related: https://aviation.stackexchange.com/a/95531/34686

@CharlesBretana  "G" stands for "gravity" - 1G = 1.0 x the earth's gravity, 2 G's = 2x earth's gravity, and so on. Gravity is a force. That force can produce an acceleration, but gravity is a force, and a "G" is that force, normalized by dividing out the mass of whatever is being discussed (pilot, aircraft, rocket payload, etc). At 1G, there is no "delta V" happening. That one G is not "an acceleration".

In normal Pilot Math & Engineering, gravity is a force, and it can cause an acceleration. If I drop an object, it accelerates down; that change in velocity is described by F=MA, and gravity acts like every other "force" we encounter.  Relativistic physics & unobservable frames of reference and abstracting away "the force of gravity" is all well & good, but when you're discussing aviation with actual aviators, gravity most certainly IS a force that is recognized and understood, so that it can be safely overcome with lift and thrust and lots & lots of money.

Aviation works and is taught & understood in Newtonian physics. Besides, the question is about terminology & convention, not about physics.  Injecting "curvature of space-time" into this discussion is unhelpful & unproductive to understanding why the G-meter points to "1" rather than "0" when at rest.

It really depends on the convention. I've seen both answers used before, but most Avionics software I've seen assumes the 0g convention so that normal acceleration is a rough approximation for the amount of vertical acceleration.
ARINC for example defines multiple labels for IRUs.  The standard one is Label 333 for Body Normal Acceleration (0 g on ground), but there's also for on some hardware label 370 unbiased normal acceleration (1 g on ground).
It seems intuitive to have normal acceleration be 0 when in constant-speed straight and level flight as this translates easily to other parameters like vertical acceleration and works better for math based on ratios and magnitudes. For example with this convention 0.75 g is just as bad as -0.75 g. On the other hand having normal acceleration typically be 1g makes it easier to translate into lift or load factors like Az/W. Whichever way, this bias to 0g usually doesn't account for roll or high pitch- the bias to normal acceleration should really be cos(pitch)*cos(roll)*(1g). So no matter the convention you'll likely need to subtract the gravity vector and there's no free lunch.

Wow, can terminology produce some lively discussion. No wonder the world has so many languages!
Here is a reference on where the term "normal acceleration" comes from and how can apply to the z axis of aircraft.  Normal acceleration is actually a component of centripetal force perpendicular to the path of an object moving in a circle.
Tangential acceleration is a component of force acting in the path of the object affecting its velocity (operating against drag).
With aircraft, in straight and level flight, the "normal acceleration" of 1 G along the z axis (approximately)$^1$ balances the gravitational force of 1 G.
Sitting stationary on the ground, normal acceleration of the aircraft is 0.  The aircraft is not producing any force.
Why we have G meters?  They are for the human limits as well as the structural limits.  It is important to understand the difference due to gravity between the "felt" loads and what the plane is actually doing.  Back to our 30 degree turn.  G forces are 1.15.  The plane is actually generating 0.57 G by accelerating to the side, with 0 vertical acceleration (gravity balances the vertical force), and is also yawing to keep the nose aligned with the relative wind as much as possible (lowest drag).
The "ball" is also accelerating sideways at 0.57 G, but because it has no vertical acceleration, it also has 1 G of gravitational "pull" downwards.  As a result, the ball will align itself with the z axis as long as the nose is kept coordinated.
If the nose is not properly aligned, then the glass tube/ball are no longer in the plane of the "normal force".  This will show as the ball being too far outside in a skidding turn, and too far inside in a slipping turn.
$^1$ the vertical component of it, taking angle of attack into account

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What ceiling does the FAA consider to be controlling for FAR 91.155(c)?
What ceiling does the FAA consider to be controlling for FAR 91.155(c)?  The ceiling reported at the airport for which the surface-level controlled airspace is designated, or the ceiling directly above the airplane?
What specific guidance has the FAA issued on this topic?
FAR 91.155(c) prohibits VFR flight (in the absence of a Special VFR clearance) "beneath the ceiling under VFR within the lateral boundaries of controlled airspace designated to the surface for an airport when the ceiling is less than 1,000 feet."
Note that in some cases surface-level controlled airspace extends more than 25 miles from the airport for which the airspace is designated.
Just for clarity, let me emphasize that this isn't meant to be a question about the status of "extensions" in relation to FAR 91.155(c), so maybe SIT/PASI (linked above) is a somewhat ambiguous case-- I just picked it because the surface-level controlled airspace extends so far from the airport.  KSHR in Sheridan WY is another case with rather expansive surface-level controlled airspace, that avoids the "extension" issue.
Or to "put a finer point on it":

If the airport (for which the surface-level controlled airspace is designated) is reporting a 900' ceiling, does that mean I can't operate VFR (without a Special VFR clearance) below any cloud ceiling, regardless of its altitude, anywhere in the surface-level controlled airspace?  Or if I deem that the ceiling in a distant portion of the surface-level controlled airspace is clearly above 1000' AGL, am I fine to operate under VFR in that portion of the airspace, below the ceiling, without a Special VFR clearance, so long as I stay 500' below the clouds and meet the other rules applicable to flight in Class E airspace?

Same as 1), but I'm operating in a distant portion of the airspace where there is no cloud ceiling at all.  Do I still need a SVFR clearance?

Conversely, if the airport (for which the surface-level controlled airspace is designated) is reporting a 2000' ceiling, but my flight path will take me over high terrain where the ceiling appears to be lower than 1000' AGL, am I ok to operate under VFR without a Special VFR clearance, as long as I can stay 500' below the clouds and meet the other rules applicable to flight in Class E airspace?

If the reported ceiling at the airport is considered to be the controlling factor for FAR 91.155(c), and automatically causes the entire surface-level controlled airspace to be considered "IMC" (i.e. off limits to VFR traffic without a Special VFR clearance) up to some level, even if flying in a distant portion of that airspace where there are no clouds, then what defines the top of that off-limits ("IMC") airspace?  Does the top of that off-limits ("IMC") airspace have a constant AGL height, equal to the reported cloud ceiling at the airport?  Or does it have a constant MSL height, equal to the MSL equivalent of the (AGL) ceiling reported at the airport?

Another way to think about the question, is to ask "when flying VFR without a Special VFR clearance, if I'm only going to be operating in a distant portion of the surface-level controlled airspace, in what circumstances am I obligated to determine the reported cloud ceiling at the airport for which the airspace is designated, via listening to the radio or other means?  Any time I think there may be a ceiling 1000' or lower over the airport?  Or only in cases where that information would help me ensure that I don't fly below a ceiling lower than 1000' AGL at my location?"

But isn’t “the ceiling directly above the airplane” also known as cloud clearance? (is there any definition of ceiling that uses the aircraft as a reference base?!)  And if the pilot is maintaining 500’ AGL or above, the weather would actually be 1500’ or better, which is VMC, which kinda renders the question moot, right? I think you can back into the correct answer here without a lot of mental gymnastics...

@MichaelHall -- I wasn't meaning to suggest measuring from the airplane up to the cloud base, it's just a question about the height of the ceiling above the ground being irregular (perhaps because ground is irregular, or because cloud is limited in extent, etc...)

There are some related questions, but most seem to be specifically asking about the "extension" issue which I'm really not trying to get back into right now.

Someone actually familiar with VFR operations at Sitka would be well positioned to answer the question.  They've actually recently reduced the airspace at Sitka.  This article https://www.aopa.org/news-and-media/all-news/2022/september/21/faa-shrinks-class-e-surface-area-around-sitka implies that the conditions at the airport were considered controlling for the whole airspace.

@757toga -- I follow your logic-- thanks-- hoping to get some answers based on actual practice--

considering removing the SVFR-tag -- this question is really about what happens if I am not flying under SVFR clearance--

The language talks about operating "beneath the ceiling ... when the ceiling is less than 1,000'."  If you're not beneath that sub-1,000' ceiling, then this language doesn't apply to you.
In your point 2, if there's no ceiling where you are, then you definitely aren't operating "beneath the ceiling."
Your point 1 sounds like the cloud layer is sloping up as you get farther from the airport, and the pilot knows (how?) that where he is, there is now more than 1,000' between the ground and the clouds, while your point 3 has the ground sloping up so that the space between the ground and the clouds has become less than 1,000' where the pilot is operating.  Those points would boil down to asking if the definition of ceiling,

Ceiling means the height above the earth's surface of the lowest layer of clouds or obscuring phenomena that is reported as broken, overcast, or obscuration, and not classified as thin or partial.

is related to the (clouds & earth's surface at the airport itself), or the (clouds & earth's surface where you're operating).  I'll leave that distinction for a separate answer.

@757toga -- yes, in fact I recall you posted a detailed comment to that effect once before.  And yet-- considering that control zones had no defined upper limit-- even then, was it really correct to say that "the entire control zone was IMC"?  What if you were above any ceiling?  Food for thought...


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Am I slipping or am I skidding?
I'm in a shallow left bank, maybe 15 degrees. Right rudder pedal is all the way to the floor.  Flight path is curving toward the right.  Ball is far to the left corner of the glass tube.  Yaw string1, 2, 3 is blowing with the tail end deflected toward the right.
Am I "slipping" or am I "skidding"?
Note that ball is deflected toward outside of turn, but also toward low wingtip.
Update: I've been prompted to edit to explain why the current question is not a duplicate of What is a skidding turn (vs slipping turn)?.  That question is primarily about the influence of pitch control inputs on slip or skid.  If that question were construed to also encompass my question, it should arguably be closed for being overly broad.
Footnotes:

It's a pusher plane with no appreciable propwash over the nose, therefore the yaw string deflection is strongly correlated to the direction of the free-stream airflow

Descriptions of what a "yaw string" is -- https://www.yawstring.com/about, https://en.wikipedia.org/wiki/Yaw_string

Information about "yaw string" arguably makes no difference and would not change the answer to question.  If you go test fly this maneuver in your Cessna 152/172 etc and reply accordingly, you need not tape a yaw string on your windscreen if you'd rather not risk getting adhesive residue on the plexiglass.  (But a yaw string actually does work half-way decently even on a tractor-engine plane; there's a still a strong correlation between yaw string deflection and ball displacement.)

Is this a trick question?  You’re usually the guy providing detailed answers on these things…

Does this answer your question? What is a skidding turn (vs slipping turn)?

Then I'd recommend fixing it.  Re-asking essentially the same question from a subtly different angle doesn't declutter this site, or clarify anything.  It just adds to confusion.  We should seek to simplify.

There's also this closed one:  https://aviation.stackexchange.com/questions/97114/what-causes-a-slip-skid-turn?noredirect=1&lq=1

And this one:  https://aviation.stackexchange.com/questions/79338/what-are-the-forces-present-in-a-coordinated-turn?noredirect=1&lq=1  There are likely others, plus many, many other sources of information available to explain this simple concept.

I'm not even convinced it is a corner case.  Heading and ground track are irrelevant to the definition of a slip or skid.

Let us continue this discussion in chat.

ASE mods have frequently told me not to use comments for discussion, so as soon as the option to go to chat arises, I always take it-- see https://chat.stackexchange.com/rooms/143040/discussion-between-quiet-flyer-and-michael-hall --  all previous comments deleted--

This is a common glider maneuver called side-slip. Aileron and rudder inputs can be balanced to result in a straight flight path. Glider pilots use it to dissipate energy and burn altitude w/o increasing airspeed beyond what's possible with regular air brakes for a steep landing approach. It's a stable flight configuration in a glider. It takes some time getting used to see the runway from the side window on final.

(Let's hope this is being done at altitude after a few clearing turns).
OK!  The yaw string says slipping to the left. (Deflected towards the right).

what kind of slip?

A forward slip.  If you want to stop curving to the right, add more left aileron (or reduce right rudder).  Ailerons control flight direction in a forward slip.

other factors

The pusher prop may be adding to the stability of this configuration, but it is cross controlled.  Best done a good margin above stall speed.
PLEASE CHECK WITH AN EXPERT REGARDING SPIN ENTRY AND RECOVERY IN THIS AIRCRAFT.

@quietflyer you could say a horribly uncoordinated skid to the right, but who in their right mind would turn like that?  Massive amounts of drag anyways.

@RobertDiGiovianni --well I did it in a Challenger ultralight a couple years ago, just for fun--

Since airplanes go mostly forward, aren't all slips forward slips?

@MichaelHall Ha, well, a long time ago we talked about side slips.  This is a good example of why the forward slip is a close cousin, but the application is different.  What the OP is basically referring to is a battle between fuselage (skid) turning forces and wing (much stronger) turning forces.  In a forward slip, full (if the airplane allows) rudder deflection creates as much drag as possible, with opposite bank angle (ailerons) determining ground track.  Side slip bank angle holds against crosswind, while rudder (as needed) keeps nose straight.


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What is the name for the spherical type of attitude indicator that also shows heading?
Is there a special name for the type of attitude indicator or artificial horizon that is more or less a complete sphere and includes lines to show the aircraft heading?  I've seen them in the cockpits of military aircraft from the 1960s.
(I'm not asking about instruments specific to space flight.)
The type of instrument I'm asking about may be seen here (link to commercial photo site).

@mins -- I appreciate the links, but am not seeing the heading marks on the instrument shown there

You're right, that not the same instrument.

I have never seen such a device with a magnetic heading references integrated into it.  Given that it appears to have flight director and autopilot command bars integrated into it, that kind of an instrument generally goes by the name flight command indicator (FCI), but these devices have all kinds of different names, e.g. artificial horizon, attitude indicator, attitude reference indicator, etc.
Now, according to the cockpit layout for the F-4 fighter, the instrument appears to be called an Attitude Direction Indicator (ADI).

ADI, for Attitude Direction Indicator.

This wikipedia article shows an ADI without the heading marks the OP references. That matches the terminology I'm familiar with; we refer to "the ADI" portion of the Primary Flight Display (the part with blue sky & brown ground), and ours doesn't have heading marks on the horizon line either.

Sure thing, partner! In the aviation world, we use a fancy-sounding term called "attitude indicator" to refer to a type of instrument that helps us pilots keep the airplane level and flying straight. Now, there's a special kind of attitude indicator that's shaped like a sphere and also tells us which direction we're heading. We call this little gizmo a "spherical attitude indicator with heading." Pretty nifty, ain't it?

If you can provide a link to an external source using this terminology, I'll upvote and accept this answer!


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Pitching up with constant bank -- how to compute final attitude, from initial attitude and degrees of rotation about pitch axis? (Specific example)
(Note: highly related, but different: How to compute final aircraft attitude, if we know starting attitude and degrees of rotation around each axis? (Specific example))
I'm flying due north (000 degrees heading), with bank angle of 0 degrees and a pitch attitude of 0 degrees.
I roll the airplane 30 degrees to the right, rotating only around the aircraft's roll axis (longitudinal axis).
Then I pitch the aircraft up through 45 degrees of rotation around the aircraft's pitch axis (lateral axis), with no rotation around the airplane's yaw axis (sometimes called the "directional" axis), but rotating as needed around the aircraft's roll axis (longitudinal axis) to hold the bank angle constant.
At the instant the pitch rotation is finished, what is the airplane's heading and pitch attitude?  And what are the relevant formulae?
Note that -- just as in the related question linked above -- no information is given about the instantaneous direction of the flight path at any point in the maneuver.  We could imagine that the angle-of-attack remains constant, in which case large variations in thrust and in airspeed would be required to fly the maneuver.  Or we could imagine that the maneuver involves a very large change in angle-of-attack and minimal change in the direction of the flight path, at least as seen in side view, resembling some sort of modified version of the famous "Cobra" maneuver.  Or we could imagine that the maneuver is being flown by a hovering "Harrier"-type VTOL airplane, or a helicopter, with zero airspeed.  Or we could imagine the maneuver being flown in some hypothetical way that would never be possible in reality (at least in a fixed-wing aircraft)-- e.g. the flight path is completely horizontal and linear throughout the maneuver, with substantial airspeed, and during the maneuver the aircraft develops an extreme sideslip ("yaw") angle as well as an extreme angle-of-attack.  Those details shouldn't affect the answer -- it appears that all these cases boil down to the same problem in spherical geometry.
In fact, it might not actually be possible to fly this maneuver at all in any conventional fixed-wing aircraft in a way that exactly meets all the constraints of the problem, including zero rotation about the yaw axis.  It appears that it would be possible to fly the maneuver in some helicopters, or a Harrier-type VTOL aircraft operating with zero airspeed, or in some as-yet-unbuilt future derivations of such aircraft.  The point being, this is not a question about the flight dynamics of any particular aircraft.  It's a question about how to compute the final bank angle, final pitch attitude, and delta heading, from a known initial attitude and a prescribed series of rotations about the various axes of the aircraft.

Wait, did you not a already get an answer to this question last time? Is this not a duplicate? Or what is the difference to your last question? Is it the part in bold?

You'd end up at the same pitch attitude as the other question, but you would need to know roll and pitch rates to calculate heading.

@U_flow -- yes, the bold italicized portion makes it a different question which I believe will have a different answer.

@MichaelHall -- re your comment-- I don't believe that's the case.  I believe the geometry is "locked in" in such a way that the same delta heading will result regardless of the rate at which the pitch and roll rotations are carried out.  Note that the specification of zero rotation about the yaw axis means that we can't imagine that slowing down the pitch rate to a very low value would mean that we would end up spending a lot of time flying something that closely resembles a normal climbing turn -- if that were the case, then we'd need a lot more information to solve the problem.

I don't think so.  Try it sometime, snap roll to 30 deg AOB - you can get there with virtually no heading change.  Now take a full two minutes to slow roll to 30 - you are turning the entire time you are rolling.

My point is that you need to factor in the roll rate to compute heading.

Let us continue this discussion in chat.

Going to chat re conversation w/ M.H.-- and deleting most of comments-- anyone else wishing to chime in might as well use this room too rather than creating a new one -- https://chat.stackexchange.com/rooms/144788/discussion-between-quiet-flyer-and-michael-hall --

@quietflyer in chat:  "But the question as I see it is essentially a problem in three-dimensional spherical geometry, not flight dynamics."

Given that this is purely a 3D geometry puzzle that does not involve flight dynamics, I'd recommend removing the phrases "I'm flying due north", "I roll the airplane", and "Then I pitch the aircraft" so you don't lead anyone else down the rabbit hole that we've explored together. Replace with something like: "imagine I fixture a pointing device in a gimbal and then..."  And remove the "flight dynamics" tag!  DV for being so misleading.

Comments immediately above are addressed to some extent in chat --

@quietflyer Please do not "bump" your question with minor edits.

Just as in the related answer How to compute final aircraft attitude, if we know starting attitude and degrees of rotation around each axis? (Specific example), this answer will use a navigational computation to solve this problem in spherical geometry.
We'll imagine the aircraft to be centered in the middle of an imaginary globe with the nose pointing at 0 degrees latitude and 0 degrees longitude.  We'll set the specified total degrees of rotation about the aircraft's pitch axis (45 degrees), equal to the total angular path length that would be followed by a vessel starting at 0 degrees N 0 degrees E, and travelling along a rhumb line, which is a line of constant heading.
The aircraft's bank angle-- 30 degrees--will be represented by the vessel's heading.  In this problem the bank angle is constrained to remain constant-- that's why we are modelling the problem with a rhumb line calculation rather than a great circle calculation.  Note that maintaining a constant bank angle requires some rotation about the aircraft's roll axis --there must be a non-zero rate of roll toward the high wingtip.
The final latitude of the vessel will represent the aircraft's final pitch attitude, and the final longitude of the vessel will represent the aircraft's total heading change.
The particular rhumb line calculator used to solve this problem gives the option to use a "normal sphere" for the shape of the earth, in which case the expected input for distance travelled (path length) is in "geographical miles".  A geographical mile is very nearly equal to a nautical mile, but is defined to be equal to exactly 1 minute of longitude at the equator, so for the distance travelled, we'll multiply the specified degrees of rotation about the aircraft's pitch axis (45) by 60 minutes / degree to get the total angular path length in units of minutes of equatorial longitude, which is also the total path length in "geographical miles".  (The value is 2700.)
The calculator outputs a final position of 38 degrees 58.269 min N, 24 degrees 19.530 minutes E.
Converted to decimal degrees, this means that the aircraft's final pitch attitude is 38.97 degrees, and the aircraft's change in heading is 24.33 degrees.
Note that this is about 2.2 degrees greater change in pitch attitude, and 2.2 degrees less change in heading, than in the related problem where we had the same 45 degrees of rotation about the aircraft's pitch axis, but rather than constraining the bank angle to remain constant (which requires some rotation about the aircraft's roll axis), we specified that there was no rotation about the aircraft's roll axis (which led to a 9.1 degree increase in bank angle during the course of the pitch rotation.)
It's perhaps a little surprising that there wasn't a greater difference in final results between the two related problems.  Pitch rotation values of 60 degrees or more would likely yield a much greater difference in final results between the two related problems.

Also it is the case that March 23 edits to the question were specifically in response to feedback from another ASE user.

I've prepared a table of results specifically addressing the last sentence of question-- which is rather interesting-- showing the results of pitch rotations of 60 though 90 degrees--

It should be possible for others to answer this question. If you believe you have an answer that is better than the present one, consider addressing a moderator (perhaps via a "@username" comment right here) with a request to temporarily re-open. It's kind of silly for an ASE question to have only one answer, which is heavily downvoted.


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Would twin-engine airplanes be safer (in case of engine failure) if had minimal dihedral?
In the specific context of the sudden failure of one engine, would twin-engine airplanes (with wing-mounted engines) be safer if designed with minimal dihedral, so that aerodynamic coupling between sideslip and roll torque was minimized?  (In the case of swept-wing aircraft, or high-wing aircraft, that might actually require some anhedral (like the Tupolev Tu-104).
If not, why not?
If a specific example is needed, then take for instance the Beechcraft King Air, but the question is not intended to be limited to that.  Swept-wing designs are also of interest.

OT but I was typing an answer for the guy that posted the Piper Cub question before it was deleted. What he saw was a Cub with a left side door mod, which incorporates the throttle into the same spot it was without the door, using a mechanism to transfer its movement to the throttle linkage when the door is closed. Left side doors are done on Cubs, and Citabrias/Champs, for operating on floats so you can get out either side. It becomes a big deal when you can only approach a dock from one side due to winds or space constraints. I owned a Citabria on floats with the single right side door.

@JohnK-- thanks, was wondering where that question went-- kind of a shame, that would have been a worthwhile answer-

I've created a chat room intended for discussion of all aspects of the current question, including any answer-- I've copied one comment chain under another answer over there already-- https://chat.stackexchange.com/rooms/144335/a-room-to-talk-about-answers-to-question-would-twin-engine-airplanes-be-safer-i

When flying single engine in a twin, the normal technique is to apply ~5 deg of bank into the live engine. This actually cancels dihedral effect because the offsetting force of the rudder, being applied to keep from turning, prevents the plane from sideslipping toward the low wing, and without sideslip, dihedral effect can't work.
You are actually flying along straight through the air, that is, no sideslip, and if there is a yaw string on the windshield it will be straight, even though the wing of the live engine is low because the rudder application is "holding the plane up" on an inclined plane (the bank) you might say.
If you don't apply the bank into the live engine, the lateral rudder force being applied to keep the nose pointed straight results in the plane sideslipping toward the dead engine with wings level. Dihedral effect will create a rolling moment the opposite way, toward the live engine, and to hold the wings level you will actually be holding a small amount of aileron into the dead engine as you slew along. Bad news all around.
So the proper steady state coordinated flight condition for a straight wing twin is non-side slipping flight, which means dihedral effect is not even a factor, while slightly banked. This is why when you are correctly configured, the ball will be pointing at the ground, which because you are banked a bit, means it's offset in the tube by the amount of bank, say half a ball width.
So the result is you still design for the dihedral effect you want for desirable lateral stability in normal flight.
In fast swept wing airplanes (at least in the CRJ - guys here who fly heavies with wing mounted engines may have something to say) you didn't try to lower the wing into live engine; you just kept wings level and the brick (skid ball) centered. I was never able to find out exactly why banking into the live engine wasn't necessary, but I think that with the thrust reserve you have with a jet while single engine, it wasn't worth the effort to hold the bank because the moderate sideslip you get with tail mounted engines wasn't worth correcting for.

The combination of offset thrust forward, and lateral thrust applied by the rudder to prevent a yaw into the dead engine, results in a resultant thrust vector that is, say, 10 deg into the dead engine with wings level. Banking into the live engine tilts the lift vector, introducing an offsetting lateral thrust vector. If the thrust was symmetrical, the tilt initiates sideslip, which the fin coverts into a turn. The thrust being asymmetrical, with a lateral thrust being applied by the rudder, the sideslip tendency offsets the rudder's lateral force, and you fly straight even though banked.

The instinct during multi-engine training is, engine quits, you apply rudder to stop the yaw, and bank 5 deg into the rudder application. It becomes automatic. You glance as the ball, and it should be sitting offset in the vial appropriate to the 5 deg bank angle, indicating no lateral acceleration is happening.  5 deg is a ballpark number that gets you close enough; the precise value for a given airplane may be 4, or 6, or whatever. If the plane has a windshield yaw string, you just straighten it. Doing this correctly can mean the difference between climbing and descending in a piston twin.

That makes sense--

I appreciate the reply to the lengthy comments-- I think there are some subtle points here about what we are really accomplishing by slightly banking the aircraft toward the good engine-- copying all comments to chat room https://chat.stackexchange.com/rooms/144335/a-room-to-talk-about-answers-to-question-would-twin-engine-airplanes-be-safer-i, feel free to post any more thoughts-- thanks

For a GA twin with one failed engine, the difficulty one always hears about is yaw authority: can the fin and rudder at least prevent the airplane from circling?
Yes, if sideslip is needed to maintain course, less dihedral means less roll.  But aileron authority isn't reduced with one engine out, so correcting for roll should be easy.  Certainly less of a safety concern (even within the realm of pilot workload) than managing the airplane's reduced performance and finding where to land.
(John K mentions examples where quite a bit of aileron is needed.)
Over all the durations of all the flights made by the whole fleet of a type, overall safety is likely optimized by choosing the dihedral that's best for the most common flight regime: both engines running.

Here it is important to differentiate between wing "dihedral", which is there for roll stability, and "dihedral effect", which (including wing dihedral and all side area above the CG$^1$) affects roll tendency in a slip.

removing wing dihedral is probably the worst way to reduce dihedral effects

Now, let's see what a 5 degree bank angle does with a twin engine aircraft losing one engine.

banking the plane causes it to slip to the side it is banking.

This has a beneficial effect of creating side pressure on the tail: a yaw torque into the live engine.  This enables rudder input to be reduced.

what about heading?

We actually need to create a forward slip to maintain heading.  Very safe, stable configuration, but, as John K pointed out, not the best for long distance due to higher drag.  Perhaps better (depending on model) to use rudder to compensate for live engine yaw and hold fuselage straight for fuel economy, at the expense of reduced available rudder travel into the live engine (for turning)$^3$.

Would reducing wing dihedral make it "safer"?

From the standpoint of rolling tendency in a slip vs available roll control, a better approach much be to increase roll control authority with larger ailerons and/or spoilers, and to seek, in design, to minimize excess dihedral effect$^4$.
It was heartening to learn the Tu-104 also suffered Dutch Roll, even with wing anhedral$^2$.
$^1$ items that create greater (lateral) yawing and rolling force, such as T-tails, large and tall vertical stabilizers, and fillets must be carefully considered.
$^2$ see the more "traditional" Tupolev Tu-114
$^3$ maintaining adequate airspeed is critical
$^4$ see B-52 Stratofortress

@quietflyer generally, if it banks that way its gonna move that way.  But thanks for creating chat. I will check it out.

Banking HAS to create a slipping tendency, because banking tilts the lift vector. The slipping is the lateral movement in reaction. If airplanes didn't have vertical fins, they would turn by sliding sideways, like a helicopter moving sideways in hover when the rotor disc is tilted. The "turn" only happens because the fin keeps the body weathervaned into the stream as it moves laterally, resulting in an arc.

I've copied all comments above over to the chat room https://chat.stackexchange.com/rooms/144335/a-room-to-talk-about-answers-to-question-would-twin-engine-airplanes-be-safer-i, and deleted my own.  Among other things they address whether or not banking really creates a significant slipping tendency in general, versus in the specific case where the pilot is using the rudder as needed to hold heading constant. -- further discussion there is welcome-- Robert DiGiovanni has also posted some new content there.


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How to compute final aircraft attitude, if we know starting attitude and degrees of rotation around each axis? (Specific example)
I'm flying due north (000 degrees heading), with bank angle of 0 degrees and a pitch attitude of 0 degrees.
I roll the airplane 30 degrees to the right, rotating only around the aircraft's roll axis (longitudinal axis).
Then I pitch the aircraft up through 45 degrees of rotation around the aircraft's pitch axis (lateral axis), with no rotation around the airplane's roll axis (longitudinal axis) or yaw axis (sometimes called the "directional" axis).
At the instant the pitch rotation is finished, what is the airplane's heading, bank angle ("roll attitude"), and pitch attitude?  And how is it calculated?  Is calculus involved?
It's obviously a problem in spherical geometry...
Context: If the answer seems too obvious, consider what happens if we rotate 90 degrees around the pitch axis (longitudinal axis), starting from 0 degrees pitch attitude and a 90-degree bank angle.
Admittedly, the maneuver described in this question is a bit hypothetical-- it might not be possible to actually fly this maneuver.  It seems to describe sort of a chandelle, but any actual climbing turn is generally going to involve continual rotations around all three axes.  But, humour me and imagine that it somehow is possible to fly this maneuver.

Closely related:  What is the relation between roll angle and pitch angle?

Have a look at the rotation matrices in this answer.

@Bianfable -- it's not clear to me that that answer really tells me how to compute the final bank angle, pitch attitude, and heading, after the prescribed rotations.  Rather, that answer converts a vector from body coordinates to "navigational" coordinates.  Maybe if I understood the situation better I would see how the two problems are essentially the same, but I'm not seeing it at the moment--

Anyway, I've made chat room for the present question-- https://chat.stackexchange.com/rooms/144514/a-room-to-talk-about-answers-to-question-how-to-compute-final-aircraft-attitude .  Feel free to chime in.

Related-- https://aviation.stackexchange.com/questions/66945/what-are-the-exact-meanings-of-roll-pitch-and-yaw/66955#66955.  -- in that answer to another question I asserted that if the aircraft is banked, rotating about the aircraft's pitch axis will change the bank angle even with no rotation about the aircraft's roll axis or yaw axis.It's challenging to discern whether or not any existing answers take this into account

Highly related but different -- https://aviation.stackexchange.com/q/98210/34686

Bear in mind that you usually don't compute attitude changes on their own. They are one part of the puzzle in a flight simulation. The depth and fidelity of this simulation is up to you. To better understand the intent behind your question: what exactly are you working on or thinking about?

@Flo -- it's really not a "simulation".  I believe that the question as currently written contains all the information needed to fully constrain the problem so that there is only one solution.  As far as "what I am thinking about", perhaps the last two paragraphs in this related question https://aviation.stackexchange.com/questions/98210/pitching-up-with-constant-bank-how-to-compute-final-attitude-from-initial-at may shed some light on the question.  (ctd)

(ctd) The only difference between the two questions, is that in one question the rate of rotation about the roll axis is constrained to be zero, while in the other question the bank angle is constrained to be constant.

(ctd) Note that at least one ASE user reacted  negatively to the other related question, essentially saying that I should have made it more clear from the start that it wasn't really an aviation-related question!  But, I think that it actually is...

First off: for your specific example, no calculus is required (but quite a bit of linear algebra), since you only rotate around one axis at a time. If you start doing complex maneuvers, combining different rotations at different rates and times, you will need calculus (in the form of numerical integration aka "simulation": basically doing everything below in very small steps).
Rotations come in two 'flavours': intrinsic and extrinsic. For intrinsic rotations, the coordinate system rotates with the object. So much like inside an airplane. For extrinsic rotations, you rotate around fixed axes of the coordinate system, but that's not relevant for your question.
These simple rotations can be described most intuitively by rotation matrices. You start with some vector, and premultiply it by these rotation matrices to get a new vector in a new direction. For example, let's consider an airplane moving in the x-direction, wings parallel to the y-direction, and the roof pointing in the positive z-direction. The direction of the nose is then $\begin{bmatrix}1 & 0 & 0 \end{bmatrix}^T$, the direction of the left wing $\begin{bmatrix}0 & 1 & 0 \end{bmatrix}^T$ and the direction of the ceiling $\begin{bmatrix}0 & 0 & 1 \end{bmatrix}^T$. In principle, two of these vectors uniquely describe the orientation of the plane, but let's keep all three for completeness and put them together in a matrix describing the original orientation of the plane: $$\mathbf{O} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$ Now, we want to rotate this around the roll axis by -30° and pitch axis by -45° (negative because of the right-handed coordinate system used): $$\mathbf{R_{roll}}=\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos{-30°}& -\sin{-30°} \\ 0 & \sin{-30°} & \cos{-30°}\end{bmatrix} \qquad \mathbf{R_{pitch}}=\begin{bmatrix}\cos{-45°} & 0 & \sin{-45°} \\0 & 1 & 0 \\-\sin{-45°} & 0 &\cos{-45°}\end{bmatrix}$$
The order of multiplication matters here, and in this case, we must do the rotations from left to right$^*$, even though the matrix multiplication goes from right to left, with $\mathbf{F}$ the final orientation
$$\mathbf{F}=\mathbf{R_{roll}R_{pitch}O}$$
We find the final orientation:
$$ \mathbf{F}=\begin{bmatrix}
\sqrt{2}/2 & 0 & -\sqrt{2}/2 \\ \sqrt{2}/4 & \sqrt{3}/2 & \sqrt{2}/4 \\ \sqrt{6}/4 & -1/2 & \sqrt{6}/4
\end{bmatrix}\approx \begin{bmatrix}0.7 & 0 & -0.7 \\ 0.35 & 0.87 & 0.35 \\ 0.61 & -0.5 & 0.61\end{bmatrix}$$
Remember: the columns of $\mathbf{F}$ were nose, left wing, ceiling, in that order. So basically, the nose points a bit to the left and a bit more forwards than up; the left wing still points at the same heading as before but a bit down; and the ceiling points backwards and quite a bit more up than left. Got it? :) actual values coming soon.
$^*$ Really, we first pre-multiply by the roll, and then pre-multiply by the pitch transformed to the rolled coordinate system, so really $(\mathbf{R_{roll}R_{pitch}R_{roll}^T})\mathbf{R_{roll}}$ But that's just unwieldy, don't you think.

Regarding the parts in your question about the 90° rotations: the parameters pitch, yaw and roll are ill defined for extreme attitudes so it's sometimes hard to go back from the F matrix to these parameters. But I think by using the full matrix O at least we prevented any gimbal lock.

Question: (and this would apply to the other answer too)-- is the final bank angle really 30 degrees?  For any given initial bank angle, if I rotate far enough (in the nose-up sense) about the aircraft's pitch axis, with no rotation about the yaw or roll axes, don't I increase the bank angle, eventually reaching 90 degrees bank if I rotate far enough?

I'm just "flying" it with my hand ("pivoting" my hand about "pitch" only, starting with non-zero "bank") when I reach that conclusion--

I made chat room for this question-- https://chat.stackexchange.com/rooms/144514/a-room-to-talk-about-answers-to-question-how-to-compute-final-aircraft-attitude.  Have primarily focused on diagram in other answer but feel free to chime in with any more input re your answer.  Perhaps of particular relevance -- https://chat.stackexchange.com/transcript/message/63153825#63153825

Yeah I think I have my calculation from vector to roll angle wrong. Will fix :)

To the nearest tenth of a degree--
The final pitch attitude is 37.8 degrees.
The final bank angle is 39.1 degrees.
The heading change is 26.6 degrees.
How the results were obtained: thinking about how a spherical attitude indicator of the type described in this question would move during the maneuver, I realized that portion of the problem involving a rotation about the aircraft's pitch axis only, was analogous to a navigational problem involving a great circle route.1
So I found an on-line great-circle calculator.
I entered "0 degrees" for the initial latitude and longitude.
I played around with the final latitude and longitude until I had a value of 30 degrees for the initial heading, and a value of (45/360) of the earth's circumference, or 3110 statue miles2, for the spherical distance.
The final latitude is the final pitch attitude, and the final longitude is the change in heading, and the final course is the final bank angle.
For anyone wanting to replicate these results, the exact values I entered for the final latitude and longitude were 37.77 degrees N and 36.57 degrees E.
Since I haven't reversed-engineered the great circle calculator, I can't state the actual formulae used in the calculator.  However, I've noticed that the results are identical to those given by the formulae in another answer (this one).  I independently derived formulae based on geometry similar to (actually a mirror-image of)3 the geometry illustrated in that answer, and came up with--
delta pitch attitude = arcsin (sin (pitch rotation angle) * cosin (initial bank angle))
delta heading =  arctan (tan (pitch rotation angle) * sin (initial bank angle))
(no formula derived for final bank angle)4
These formulae give the same results as those given by the formulae given in that answer, and also as those given by the great circle calculator.
Footnotes:

Readers trying to understand the how the great circle problem relates to the current question, might find it helpful to imagine the aircraft being located inside of a transparent sphere, with a beam of light shining forward to trace a path on the surface of the sphere as the aircraft changes attitude.  The movement of the resulting dot of light could either represent the movement of the "pipper" or index dot in relation to the surface of a (spherical) attitude indicator, or could represent the movement of a vessel along a great circle route.

These calculations use a value of 24,880 statute miles for the earth's circumference, which is the average of the circumference around the poles and the circumference around the equator.  According to Wikipedia, "Earth's circumference is the distance around Earth. Measured around the Equator, it is 40,075.017 km (24,901.461 mi). Measured around the poles, the circumference is 40,007.863 km (24,859.734 mi)."

In my diagrams, the pitch angle was upward above the horizon, not downward below the horizon.  But since the diagrams are mirror-images of each other, the resulting formulae are unchanged.  Note that which leg of the triangle should be taken to represent the "flight path", depends on whether the aircraft is envisioned to be flying at a constant altitude with a non-zero angle-of-attack (as measured from the fuselage or longitudinal axis), or is envisioned to be rotating about the pitch axis and climbing while maintaining a zero angle-of-attack (as measured from the fuselage or longitudinal axis).  The same basic geometry applies in either case.

As I was doing the derivation, I had the impression that the geometry modeled by these formulae assumed that the aircraft rolled as needed to hold the bank angle constant at the initial bank angle as the aircraft rotated about the pitch axis, but it now appears that that is not actually the case.

I'm confused as to why these results appear to be the same as those given by this answer https://aviation.stackexchange.com/a/98017/34686, which I thought was built around a model that assumed a constant bank angle during the pitch rotation, which would imply a non-zero rate of roll rotation.  Clearly I am missing something.  Discuss further in the chat room here -- https://chat.stackexchange.com/rooms/144514/a-room-to-talk-about-answers-to-question-how-to-compute-final-aircraft-attitude

For the present purposes, it would be really nice to find an on-line great circle calculator that lets you input the initial position, initial heading (course), and distance travelled, and outputs the final location. Since this has no practical application for navigation, most are not set up this way-- please let me know via comment or chat post if you find one!

Naturally, googling "great circle calculator" will give some insight into the formulae involved--

Since the other answer referenced here has now been deleted, readers wishing to see the triangular geometry involved may wish to see figure 11-5 referenced in this chat post--https://chat.stackexchange.com/transcript/message/63155332#63155332


text:
When did this content first appear in the "Airspace" entry of the "Legend" section of the Airport/ Facilities directory?
When did language like

NOTE: AIRSPACE SVC TIMES INCLUDE ALL ASSOCIATED ARRIVAL EXTENSIONS.
Surface area arrival extensions for instrument approach procedures
become part of the primary core surface area. These extensions may be
either Class D or Class E airspace and are effective concurrent with....

(ital added)
or

NOTE: AIRSPACE SVC TIMES INCLUDE ALL ASSOCIATED ARRIVAL EXTENSIONS.
Arrival extensions for instrument approach procedures become part of
the primary core surface area. These extensions may be either Class D
or Class E airspace and are effective concurrent with...

(ital added)
first appear in the "AIRSPACE" item of the Legend section near the front of the "Airport / Facilities Directory" (now "Chart Supplement") published semiannually by the FAA?
(For a current example of similar text, see page 26 of the January 27 2022 issue of the Chart Supplement for the "Northeast" region -- link to PDF here.)
Edit-- the intent is not simply to find language establishing that the effective times of the "extensions" are simultaneous with the effective times of the adjoined core surface area, but rather to find examples of statements that the extensions become part of the adjoined core surface area.  Examples of this language from publications other than the Airport / Facilities Directory are also of interest, so long as they predate the earliest known example of this language in the Airport / Facilities Directory.  Similar language appears in the Aeronautical Information Manual (section 3-6-(e)(2)), but wasn't added until the 2016 edition.

Content stating that the surface-level Class E (or D) extensions ("arrival extensions") "become part of" the adjoined core surface-level controlled airspace ("primary core surface area") whenever the core surface-level controlled airspace is in effect, first appeared in the issues of the Airport / Facilities Directory effective beginning May 25, 1995.
The exact text in the volume for the Northeastern region for this date reads "NOTE: AIRSPACE SVC EFF TIMES INCLUDE ALL ASSOCIATED EXTENSIONS. Arrival extensions for instrument approach procedures become part of the primary core surface area. These extensions may be either Class D or Class E airspace and are effective concurrent with the times of the primary core surface area.  See CLASS AIRSPACE in the Airman's Information Manual for further details"
The last sentence apparently contains a typo (missing letter/s) and was likely intended to read "See CLASS E AIRSPACE..." or "See CLASS D/E AIRSPACE..."  (Note that this apparent typo was still present as late as 2001 -- see https://aviation.stackexchange.com/a/91352/34686 )
The previous volumes of the A/FD, effective beginning March 30, 1995, contain no content pertaining to these surface-level Class E (or D) "extensions".
Scans of the relevant pages of the Northeastern region volumes effective beginning March 30 1995 and May 25 1995 are attached below:

This resource may be viewed at the National Archives in College Park MD -- see https://catalog.archives.gov/id/2068505
The fact that this note appeared in the Airport/ Facility directory less than two years after the September 16 1993 "alphabet" airspace re-designation -- when the surface-level Class E "extensions" were first introduced-- would seem to shed some light on the FAA's original intentions as to whether any FARs should ever be considered to apply to the surface-level Class E "extensions" but not the core "surface areas", or vice versa.  This would seem to have some bearing on the following ASE questions:
Does FAR 91.155c apply to class E surface extensions?  (FAR 91.155c prohibits VFR operations below a 1000' cloud ceiling in certain airspace, unless conducted under a Special VFR (SVFR) clearance.)
Does an SVFR clearance extend to Echo surface extensions?
In the US, in actual practice, workload permitting, will ARTC facilities grant SVFR clearance for surface-level Class E "extensions" (E4 airspace)?
Which parts of class E airspace can an ultralight (part 103) fly in without prior ATC authorization?
What is the primary reason that the non-towered airports ACV, TVL, SGU, BIH, and SIT/PASI have E4 extensions to E2 airspace?
What indication has the FAA given that phrases like "surface area of Class E airspace designated for an airport" do or don't include E4 "extensions"?
This content in the AF/D (now Chart Supplement) would also appear to bear on the question of whether or not a Small Unmanned Aircraft System (e.g. model airplane, quadcopter, "drone") may be operated in the surface-level Class E "extensions" with no prior authorization, under the terms of either FAR Part 107 (specifically 107.41), or 49 U.S. Code § 44809 - "Exception for limited recreational operations of unmanned aircraft".  Note that the "extensions" are not currently included in the "gridded airspace" indicating the areas where prior authorization is required for SUAS operations on the LAANC map for automated on-line authorization of SUAS operations.  Several FAA issuances including the Gardner memo (scroll down to "Text of the FAA Memorandum Clarifying Class E Airspace"), and this power-point document from an FAA Aviation Safety Inspector, take the viewpoint that prior authorization is indeed not required to operate a SUAS in surface-level Class E "extensions", under the terms of either FAR Part 107 or the "Recreational Exception", even though such authorization is unquestionably required in the adjoined core Class E or Class D "surface areas" that include the airport itself.  This would seem to be inconsistent with the note in the A/F D (now Chart Supplement) that is the subject of the present answer.
The best (and most widely accepted) interpretation of the regulations appears to be that whenever the "extensions" are in effect, then from a regulatory standpoint they should be treated just like a core Class E "surface area" that actually includes the airport for which it is designated.  In the specific context of a Special VFR clearance, the most conservative approach would be to assume that FAR 91.155c's prohibition on flying VFR below a cloud ceiling that is lower than 1000' AGL (without a Special VFR clearance) does include surface-level Class E "extensions", but at the same time always get explicit confirmation from the issuing controller before assuming that any given Special VFR clearance should in fact be understood to include a surface-level Class E "extension".1
Footnotes:

The lack of a universally shared viewpoint on this issue throughout the FAA, and the resulting need for pilots to clarify with the issuing controller or controlling agency whether or not a Special VFR clearance should be understood to include surface-level Class E "extensions", was emphasized by a high-level FAA staffer during a recent (October 2021) meeting of the FAA's Aeronautical Charting Group, in discussion around a peripherally-related agenda item.  His comment included the statement "Pilots have failed checkrides due to the ambiguity around this issue."

Potential new question: in what sense do the "extensions" become part of the "primary core surface area"?  If the primary core surface area is Class D, surely no one would claim that the extensions also become Class D.  But it in light of this content in the Chart Supplement, it would seem hard to argue that any FARs that refer to a "surface area" should not also include the "extensions" (as long as the FARs do not specify a Class D surface area.)

Earliest I could find is from the Chart Supplement Pacific, 17 May 2001. Section 22 "airspace" has the phrase you seek.

@quietflyer Happy to help. What are you going to do with the info? It is kind of obscure.)

Needless to say this one is a couple of orders of magnitude better of an answer than mine :)

This is a dismal answer, but:
Instrument Rating Test Prep 2004, published 2003, contains a paragraph closely matching the one mentioned in the question: "NOTE: AIRPSACE SVC "TIMES" INCLUDE ALL ASSOCIATED EXTENSIONS..."
The partial scan found on the page I linked above matches the formatting of official Airport/Facility Directory Legend -pages.
One can thus conclude that language in questions has been used at least since 2003, and even though the document I linked is not an Airport/Facility Directory, it clearly uses one as a source/example.
Instrument Rating Test Prep 2003, published 2002, does not cite either paragraph mentioned in the question, but although the whole book is most likely scanned into the Google library project, the search may not be able to find all contents reliably. So a definite conclusion cannot be drawn that such language was not used prior to 2002.

Thanks--  examples from sources other than the A/F D would indeed be of some interest to me too, but I'm really seeking examples of language that states that the extensions "become part of" the adjoined core surface areas, not just that they are in effect at the same times.  But I've got a hunch that this is the best answer we're going to see-- have also put out some inquiries to some other pilot communities; someone may have some old back issues of the A/F D stashed somewhere--

I'm sure some of the senior members of ASE have these documets stacked in their garages, just gotta lure them here

Here is a source that establishes the use of the specific phrase you mention coincident with the 'airspace redesign'. It seems to clarify the timeframe you seek but is not in the publications you specify, so I put it here separately. The document deals with charting implications of the ambiguity you describe.

The front page of the document for reference.

Short version-- the document is a specific example of the usage of this phrase in printed materials, so that's definitely the sort of thing I'm looking for, but it doesn't happen down to narrow down the time window as to when this phrase or a similar phrase first appeared in the Chart Supplement /AFD (or any other printed material of any kind), compared to what previous answers (including your own) have already established.  It appears to me that it can't be taken as evidence that the phrase was actually used in printed materials immediately after the Sept 16 1993 airspace re-designation.

@quietflyer Sorry I didn't read all your materials. I could delete this answer unless you want the comments seen.

@quietflyer The other answer is better as it is actually helpful.) I found this looking through the congressional record.


text:
Are weather-reporting requirements for an airport in Class D airspace more stringent than for an airport in surface-level Class E airspace?
In the US, are weather-reporting requirements for an airport in Class D airspace more stringent than for an airport in surface-level Class E airspace?
Are there other requirements that would cause an airport with an operating control tower to be eligible for surface-level Class E airspace but not for Class D airspace?
The intent of the question is to understand why any airport with an operating control tower, whether "temporary" or otherwise, would have surface-level Class E airspace rather than Class D airspace during the hours that the tower was operational, if in fact this ever occurs-- apart from the obvious case of a temporary (or new) control tower going into operation in pre-existing surface-level Class E airspace.
Related ASE links: In the US, are there any operational control towers located in surface-level Class E airspace or in Class G airspace?
This outside link appears to describe the establishment of temporary surface-level Class E airspace concurrent with the establishment of a temporary control tower-- but may just be an example of inaccurate or unclear writing.  The surface-level airspace at this airport is normally Class G, and the same was true in 2016.  The only NOTAM turned up during an archive search for this airport at this time references the temporary control tower, but no temporary change in airspace designation.

thanks to ASE user randomhead for providing the link to document noted in last paragraph, in comments under another questio

Again, there's grounds for another ASE question here that focuses in on questions such as -- can airspace be temporarily changed to Class D or to surface-level Class E concurrent w/ establishing a temporary tower (but be careful not to confuse w/ part-time Class D airspace established through the usual channels for changing airspace designations, may be seasonal or even tailored to specific recurring events, as appears to have been the case at OshKosh in the past.)


text:
Are there any helicopters with tail rotors where slip-skid ball may be centered even when fuselage is streamlined to the airflow in cruising flight?
In a related answer, I stated

"The tail rotor of a helicopter also generates an aerodynamic
  sideforce. Again, just as with the twin-engine airplane with one
  failed engine, when the fuselage is streamlined to the airflow in
  forward flight, the slip-skid ball will be slightly off-center,
  deflected opposite to the direction of the tail rotor's sideforce-- at
  least in cases where the body of the helicopter tends to tilt along
  with the disk of the main rotor. If the disk of the main rotor can be
  tilted to counter the tail rotor's sideforce while the fuselage or
  body stays completely vertical, then the slip-skid ball can stay
  centered even when the fuselage or body is completely streamlined to the airflow."

I now ask:
In cruising flight in a helicopter with a tail rotor, is it ever the case that the rotor disk may be tilted to counter the tail rotor's sideforce while the fuselage or body stays completely vertical, thus allowing the slip-skid ball to stay exactly centered at the same time that the yaw string is centered, i.e. the fuselage is fully streamlined to the airflow?
Basically this is a question about how the main rotor disk system is attached to a helicopter and whether there is some way to keep the main rotor disk constantly tilted while the body stays exactly upright, in the context of a helicopter with a tail rotor.
Is this something that is possible in some helicopters with tail rotors but not others?  If so, what are some examples of helicopters where this is possible, and what is the mechanism or other design feature that makes it possible? Can the pilot somehow adjust the "tilt" (in the left and right sense) of the fuselage or body relative to the main rotor disk at will, and maintain any desired "tilt" angle for as long as he or she desires, without altering the physical relationship between the orientation of the main rotor disk and the horizon or earth?
I suppose another possibility is that the main rotor disk system is permanently attached to the fuselage at an angle that is offset (tilted to the left or right) from the fuselage to allow the fuselage to be closer to upright in cruising flight.  Is this ever done?  And is the offset ever enough to allow the slip-skid ball to be fully centered, or just closer to centered?
I realize that some (nearly all?) helicopters have either flapping or teetering blades, so maybe the angle between the disk of the main rotor system and the fuselage should not be imagined to be a definable design feature.  Still, my impression is that there is more going on than the fuselage just "hanging" freely from the main rotor-- see for example concepts in this related answer.  So it seems in general, in cruising flight, if the disk of the main rotor blade system is tilted to one side relative to the horizon in the left-right sense, the fuselage generally is too, with the direction of tilt being away from the lowered side of the rotor disk.  In other words it seems the fuselage tends to tilt relative to the horizon in the direction that reduces the amount of tilt between the fuselage and the main rotor disk, in the left-right sense, just as generally happens in the fore-and-aft-sense.  I guess I'm looking for exceptions to this rule, especially in the context of cruising flight, understanding that the disk of the main rotor must be tilted to offset the sideforce from the tail rotor.
An additional point is that even if the fuselage could be envisioned to be simply hanging freely below the disk of the main rotor blade, the sidethrust from the tail rotor would still have some tendency to tilt the fuselage relative to the main rotor disk, if the tail rotor thrust is acting significantly above or below the top of the rotor mast.  In cruising flight, it appears that the fuselage would tend to be tilted toward the low side of the main rotor disk if the tail rotor was above the top of the main rotor mast, and the fuselage would tend to be tilted toward the high side of the main rotor disk if the tail rotor was below the top of the main rotor mast.
Perhaps if the tail rotor were high enough above the top of the main rotor mast, everything could balance out and the helicopter could cruise with the fuselage perfectly upright relative to the horizon in the left-right sense, even as the main rotor disk was tilted to one side to balance out the sideforce from the tail rotor?
Restating the question once more, "Are there helicopters with tail rotors that can cruise with the fuselage perfectly level in the left-right sense at the same time that the yaw string is fully centered, and if so, what specific design features allow this?"

This answer deals with the dynamics of helicopter blade systems but not in a way that answers my specific question-- https://aviation.stackexchange.com/a/38904/34686

Yes the tail rotor's lateral thrust is offset by an opposing tilt in the main rotor that the pilot applies without realizing it, so if the tail rotor is pushing right, the main rotor will be tilted slightly left.  The machine just hangs below at whatever angle gets its C of G under the mast laterally.

@JohnK -- third-to-last and second-to-last paragraphs now may better inform the question in light of your comment.

The tail rotor also imparts a rolling torque to the airframe because on most machines the thrust line is above the body center of mass.  If you do a "bunt" in a helicopter, a zero G pushover, the tail rotor will try to roll the body under the rotor once the pendulum effect is gone.Google "mast bumping" to learn how this causes lots of grief in turbulence with teetering rotor helicopters. Articulated rotor machines just hit the blade flap stops and it's not so bad.  There are so many different forces and moments going on, with this big pendulous hunk of metal below it, I think the answer is no.

Interesting video on mast bumping - https://www.youtube.com/watch?v=_QkOpH2e6tM

In the early days of the R-22, there were numerous rotor head failures and the FAA issued an AD on the rotor, but Robinson eventually convinced the FAA that the crashes were from mast bumping, which was super easy to do in the R-22.  They are dangerous to fly in the mountains where severe turbulence can easily make an airplane go zero G and require special procedures where you try to freeze the stick while trying to re-load the rotor.  What amazes me is the US Army only figured it out in the mid 60s after the Bell 47 had been flying almost 20 years.

In steady flight (no acceleration) the slip ball position is offset by roll only.  So this question boils down to this: can a helicopter fly with 0 sideslip (streamlined) and 0 roll (wings level)?
As we know the tail rotor provides a lateral thrust force that must be countered.  The main rotor is often tilted slightly laterally to counter tail rotor thrust.  However, this will NOT result in trimmed flight with precisely 0 roll AND 0 sideslip at all speeds.  One problem is that the amount of tail rotor thrust differs with airspeed, and hence the amount of main rotor tilt + flapping would need to differ as well.  To fly (streamlined) with 0 sideslip some amount of roll will typically be required, depending on the airspeed.
 That roll will offset the slip ball, if only by a small amount (pilot's may not even notice the difference between wings level / centered slip ball and 0 sideslip).